Question

the sum of ages of 5 children born at the interval of 3years each is 50years. what is the age of the youngest child​

Answer 1

let age of first or eldest child be ¥ years

then 2nd child's age=¥+3

3rd child's age=¥+6

4th child's age=¥+9

5th or the youngest child's age=¥+12

sum of all ages=50 years

so ¥+¥+3+¥+6+¥+9+¥+12=50

5¥ + 30=50

¥=20/5

¥=4

Age of youngest child=¥+12=4+12=16years

Answer 2

Answer:

• x = 4
• Age of youngest child = 4 years.

Step-by-step explanation:

Given :

• Ages of 5 children are in interval of 3 years.
• Sum of ages = 50

To Find:-

• Age of Youngest Child.

Procedure:-

Let the age of youngest child be x

• A second child be = x + 3

• Third child be = x + 6

• Fourth child be = x + 9

• Fifth child = x + 12

Given that sum of ages is 50 years. From the mentioned information, we can infer that:-

$\sf\blue \implies\green{ (x) + (x+3) + (x+6) +(x+9) (x+12) = 50}$

Combining Like terms :-

$\sf\red \implies\orange{ (x+x+x+x)+(3+6+12)=50}$

$\sf\pink \implies \purple{4x+30=50}$

Subtract 30 from both the sides.

$\sf \green\implies \blue{5x+30-30=50-30}$

$\sf \orange\implies\red{5x = 20}$

Divide both sides by 5.

$\purple\implies \sf\pink{ \dfrac{5x}{5} = {20}{5}}$

$\blue\implies \sf \green{x=4}$

• Age of second child = x + 3 = 4 + 3 = 7 years.

• Age of third child = x + 6 = 10 years.

• Age of fourth child = x + 9 = 4 + 9 = 13 years.

• Age of fourth child = x + 12 = 4 + 12 = 16 years.

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